# Radius of convergence of $\sum_{n=1}^{\infty} \sin(\sqrt{n+1} - \sqrt{n})(x-2)^n$

by cdummie   Last Updated September 21, 2018 19:20 PM

$\sum_{n=1}^{\infty} \sin(\sqrt{n+1} - \sqrt{n})(x-2)^n$

There are two ways to determine the radius of convergence of the series:

$1. \lim_{n \rightarrow \infty}|\frac{a_n}{a_{n+1}}|$ and

$2. \lim_{n \rightarrow \infty}\sqrt[n]{a_n}$

But none of these work out easily, when i tried first one i ended up with $\frac{0}{0}$ and after using L'Hospitals rule it happened again so i couldn't find that limit, and i have no idea what could i do with the second one.

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$$\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\sim\frac1{2\sqrt n}\quad\text{as }n\to\infty.$$ Then $$\sin\bigl(\sqrt{n+1}-\sqrt{n}\bigr)\sim\sin\frac1{2\sqrt n}\sim\frac1{2\sqrt n}\quad\text{as }n\to\infty.$$

Julián Aguirre
April 06, 2016 15:34 PM

Here's a different way to tackle the problem.

At $x=-1$ the power series trivially diverges so $r\leq 1$.

At $x=1$, $$\displaystyle \sin(\sqrt{n+1} - \sqrt{n})(-1)^n=\frac{(-1)^n}{2\sqrt{n}}-\frac 18 \frac{(-1)^n}{n^{3/2}}+o\left( \frac{1}{n^{3/2}}\right)$$

Let $\displaystyle u_n=\frac{(-1)^n}{2\sqrt{n}}$ and $\displaystyle v_n=-\frac 18 \frac{(-1)^n}{n^{3/2}}+o\left( \frac{1}{n^{3/2}}\right)$.

$\sum u_n$ converges via Leibniz test and $\sum v_n$ is absolutely convergent, hence convergent (some $\epsilon-N$ work is needed to prove absolute convergence).

Therefore, the power series converges at $x=1$, hence $r\geq 1$.

Finally $r=1$.

Gabriel Romon
April 06, 2016 15:43 PM